Second My Problem First（单调队列）

最新推荐文章于 2020-07-31 19:50:38 发布

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本文介绍如何使用单调队列在O(n)的时间复杂度内求解区间最小值问题，通过实例展示如何计算给定序列中每个子区间最小值的乘积模B的值。

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Second My Problem First

 Give you three integers n, A and B.
Then we define S i = A i mod B and T i = Min{ S k | i-A <= k <= i, k >= 1}
Your task is to calculate the product of T i (1 <= i <= n) mod B.

Input
Each line will contain three integers n(1 <= n <= 10 7),A and B(1 <= A, B <= 2 31-1).
Process to end of file.
Output
For each case, output the answer in a single line.
Sample Input

Sample Output

题意：

T i = Min{ S k | i-A <= k <= i, k >= 1} ，Ti是i-A到i这个范围内S的最小值，求（T1*T2*T3…….Tn）%Bd的值。

思路:

单调队列可以在o（n）的复杂度内求出区间内的最小值，用单调队列记录区间的最小值，求出答案。

code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e7+10;
int s[maxn];
int que[maxn];
int main(){
    int n,a,b;
    while(~scanf("%d%d%d",&n,&a,&b)){
        int head = 0,tail = 0;
        ll tmp = 1,ans = 1;
        for(int i = 1; i <= n; i++){
            tmp = tmp * a % b;
            s[i] = tmp;
            while(head < tail && s[que[tail-1]] >= s[i]) tail--;
            que[tail++] = i;
            while(i - a > que[head]) head++;
            ans = ans * s[que[head]] %b;
        }
        printf("%lld\n",ans);
    }
    return 0;
}